Question

The total number of valence electrons in 4.2 g of $N_3^{-}$ ion is ($N_A$ is the Avogardo's number)

• A. $2.1N_A$
• B. $4.2N_A$
• C. $1.6N_A$
• D. $9.6N_A$

Answer 1

The correct option is C $1.6N_A$

Molecular wt of $N_3^{-}=3\times 14$=42 g

One ion of $N_3^{-}$​ has valence electrons $=3\times 5+1=16$

$42g{N}_3^{-}$ ​ $=N_A{N}_3^{-}$​ ion

valence electrons in $N_A$​ ions = $16N_A$ ​ (where $N_A$ ​ = Avogadro's number)

So, 1 g of $N_3^{-}$ ​ will have valence electrons $=16N_A/42$

therefore,4.2g of $N_3^{-}$ ​ ion will have valence electrons $=16N_A\times 4.2/42=1.6N_A$

Option C is correct.